Within the spectrum of arithmetic, graph concept which reports a mathe­ matical constitution on a collection of components with a binary relation, as a well-known self-discipline, is a relative newcomer. In contemporary 3 many years the intriguing and speedily turning out to be zone of the topic abounds with new mathematical devel­ opments and critical functions to real-world difficulties.

Assume a 1-factor F contains x. Then F cannot contain z. Then the set S = {v2, v3 , v4, v5, v6} cannot be covered by the lines of F, since S contains an odd number of points. In 1 89 1 J. Petersen showed that every cubic graph which does not have a 1-factor must have at least three bridges. He also proved that every bridgeless cubic graph is the line-disjoint union of a } factor and a 2-factor. We have, however, the following example. 22 Not every bridgeless cubic graph is 1-jactorab/e. 1 .. We shall refer to the v t v 2 v 3 v4 v� v • cycle as the outer cycle, the v 1 v9 v6 v8 Vto v1 cycle as the inner cycle, and the remaining lines as the connecting lines.

The graph on the left is the one given by the construction described above. 63 THEOREM The maximum number of edges in a connected graph of order p and external stability aoo > 3 is (p-1'+1 ) (Vijayaditya 1968a). An extremal graph can be constructed as follows. Note first that connectedness implies p � 2a00• Now take a complete graph of order p - aoo and join each of its vertices to one of a00 new ones taking care that the graph is connected. We illustrate below for the case p = 7, a00 = 3. 64 THEOREM The minimum number of edges in a graph of order p and external stability a00 is p - a00• A ny such extremal graph consists of a00 disjoint stars (including K2 and K1 as possible stars).

The first problem of this type which one generally encounters is the determination of r(K3, K3). This turns out to be 6. That 5 is too small is shown by the 5-cycle Cs. ) The remainder of this section will be devoted to examples of the above type. For various fi and Fi, graphs G of order one less than r(Fi , F; ) will be given which do not contain fi and whose complements do not contain Fi. It is of interest to note that the determination of Ramsey numbers even in the case of complete graphs is still an unsolved problem.